Problem: Let $a$ and $b$ be positive real numbers such that $a + 2b = 1.$  Find the minimum value of
\[\frac{1}{a} + \frac{1}{b}.\]
Solution: By Cauchy-Schwarz,
\[(a + 2b) \left( \frac{1}{a} + \frac{1}{b} \right) \ge (1 + \sqrt{2})^2 = 3 + 2 \sqrt{2}.\]For equality to occur, we must have $a^2 = 2b^2,$ or $a = b \sqrt{2}.$  Then $b \sqrt{2} + 2b = 1,$ or
\[b = \frac{1}{2 + \sqrt{2}} = \frac{2 - \sqrt{2}}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{2 - \sqrt{2}}{2},\]and $a = b \sqrt{2} = \frac{2 \sqrt{2} - 2}{2} = \sqrt{2} - 1.$

Hence, the minimum value is $\boxed{3 + 2 \sqrt{2}}.$